6.2 Use Lenz’s law to determine the direction of induced current in the situations described by Fig. 6.16: (a) A wire of irregular shape turning into a circular shape; 175 Reprint 2025-26 Physics (b) A circular loop being deformed into a narrow straight wire. FIGURE 6.16 6.3 A long solenoid with 15 turns per cm has a small loop of area 2.0 cm2 placed inside the solenoid normal to its axis. If the current carried by the solenoid changes steadily from 2.0 A to 4.0 A in 0.1 s, what is the induced emf in the loop while the current is changing? 6.4 A rectangular wire loop of sides 8 cm and 2 cm with a small cut is moving out of a region of uniform magnetic field of magnitude 0.3 T directed normal to the loop. What is the emf developed across the cut if the velocity of the loop is 1 cm s–1 in a direction normal to the (a) longer side, (b) shorter side of the loop? For how long does the induced voltage last in each case? 6.5 A 1.0 m long metallic rod is rotated with an angular frequency of 400 rad s–1 about an axis normal to the rod passing through its one end. The other end of the rod is in contact with a circular metallic ring. A constant and uniform magnetic field of 0.5 T parallel to the axis exists everywhere. Calculate the emf developed between the centre and the ring. 6.6 A horizontal straight wire 10 m long extending from east to west is falling with a speed of 5.0 m s–1, at right angles to the horizontal component of the earth’s magnetic field, 0.30 ´ 10–4 Wb m–2. (a) What is the instantaneous value of the emf induced in the wire? (b) What is the direction of the emf? (c) Which end of the wire is at the higher electrical potential? 6.7 Current in a circuit falls from 5.0 A to 0.0 A in 0.1 s. If an average emf of 200 V induced, give an estimate of the self-inductance of the circuit. 6.8 A pair of adjacent coils has a mutual inductance of 1.5 H. If the current in one coil changes from 0 to 20 A in 0.5 s, what is the change of flux linkage with the other coil? Reprint 2025-26 Chapter Seven ALTERNATING CURRENT 7.1 INTRODUCTION We have so far considered direct current (dc) sources and circuits with dc sources. These currents do not change direction with time. But voltages and currents that vary with time are very common. The electric mains supply in our homes and offices is a voltage that varies like a sine function with time. Such a voltage is called alternating voltage (ac voltage) and the current driven by it in a circuit is called the alternating current (ac current)*. Today, most of the electrical devices we use require ac voltage. This is mainly because most of the electrical energy sold by power companies is transmitted and distributed as alternating current. The main reason for preferring use of ac voltage over dc voltage is that ac voltages can be easily and efficiently converted from one voltage to the other by means of transformers. Further, electrical energy can also be transmitted economically over long distances. AC circuits exhibit characteristics which are exploited in many devices of daily use. For example, whenever we tune our radio to a favourite station, we are taking advantage of a special property of ac circuits – one of many that you will study in this chapter. * The phrases ac voltage and ac current are contradictory and redundant, respectively, since they mean, literally, alternating current voltage and alternating current current. Still, the abbreviation ac to designate an electrical quantity displaying simple harmonic time dependance has become so universally accepted that we follow others in its use. Further, voltage – another phrase commonly used means potential difference between two points. Reprint 2025-26 Physics 7.2 AC VOLTAGE APPLIED TO A RESISTOR Figure 7.1 shows a resistor connected to a source ε of ac voltage. The symbol for an ac source in a circuit diagram is . We consider a source which produces sinusoidally varying potential difference across its terminals. Let this potential difference, also called ac voltage, be given by v = vm sin ωt (7.1) where vm is the amplitude of the oscillating potential difference and ω is its angular frequency. Nicola Tesla (1856 – 1943) Serbian-American scientist, inventor and genius. He conceived the idea of the rotating1943) magnetic field, which is the – basis of practically all alternating current machinery, and which(1856 helped usher in the age of FIGURE 7.1 AC voltage applied to a resistor. electric power. He also invented among other To find the value of current through the resistor, we things the induction motor, ε()t = 0 (refer to Section the polyphase system of ac apply Kirchhoff’s loop rule ∑TESLA power, and the high 3.12), to the circuit shown in Fig. 7.1 to get frequency induction coil v m sin ωt = i R (the Tesla coil) used in radio and television sets and v m i = sin ωtNICOLA other electronic equipment. or R The SI unit of magnetic field is named in his honour. Since R is a constant, we can write this equation as i = i m sin ωt (7.2) where the current amplitude im is given by v m i m = (7.3) R Equation (7.3) is Ohm’s law, which for resistors, works equally well for both ac and dc voltages. The voltage across a pure resistor and the current through it, given by Eqs. (7.1) and (7.2) are plotted as a function of time in Fig. 7.2. Note, in particular that both v and i reach zero, minimum and maximum values at the same time. Clearly, the voltage and current are in phase with FIGURE 7.2 In a pure resistor, the voltage and each other. current are in phase. The We see that, like the applied voltage, the current varies minima, zero and maxima sinusoidally and has corresponding positive and negative values occur at the same during each cycle. Thus, the sum of the instantaneous current respective times. values over one complete cycle is zero, and the average current 178 is zero. The fact that the average current is zero, however, does Reprint 2025-26 Alternating Current not mean that the average power consumed is zero and that there is no dissipation of electrical energy. As you know, Joule heating is given by i2R and depends on i2 (which is always positive whether i is positive or negative) and not on i. Thus, there is Joule heating and dissipation of electrical energy when an ac current passes through a resistor. The instantaneous power dissipated in the resistor is p = i 2 R = i m2 R sin 2 ωt (7.4) The average value of p over a cycle is* p = < i 2 R > = < i m2 R sin 2 ωt > [7.5(a)] where the bar over a letter (here, p) denotes its average George Westinghouse value and <......> denotes taking average of the quantity (1846 – 1914) A leading inside the bracket. Since, i2m and R are constants, proponent of the use of p = i m2 R < sin 2 ωt > [7.5(b)] alternating current overUsing the trigonometric identity, sin2 wt = direct current. Thus, GEORGE he came into conflict 1/2 (1– cos 2wt), we have < sin2 wt > = (1/2) (1– < cos 2wt >) with Thomas Alva Edison, and since < cos2wt > = 0**, we have, an advocate of direct 2 1 current. Westinghouse < sin ωt > = was convinced that the 2 technology of alternating Thus, current was the key to 1 2 the electrical future. p = i m R [7.5(c)] He founded the famous 2 Company named after him WESTINGHOUSE To express ac power in the same form as dc power and enlisted the services (P = I2R), a special value of current is defined and used. of Nicola Tesla and It is called, root mean square (rms) or effective current other inventors in the (1846(Fig. 7.3) and is denoted by Irms or I. development of alternating current motors and – apparatus for the transmission of high tension current, pioneering 1914) in large scale lighting. FIGURE 7.3 The rms current I is related to the peak current im by I = mi / 2 = 0.707 im. 1 T F (t ) d t* The average value of a function F (t) over a period T is given by F (t ) = T ∫0 1 T 1  sin 2ωt  T 1 < cos 2ωt > = ∫ cos 2ω t dt = = [ sin 2ω T − 0 ] = 0** T 0 T  2ω  0 2ωT 179 Reprint 2025-26 Physics It is defined by 2 1 2 i m I = i = i m = 2 2 = 0.707 im (7.6) In terms of I, the average power, denoted by P is 1 2 2 P = p = i m R = I R (7.7) 2 Similarly, we define the rms voltage or effective voltage by v m V = = 0.707 vm (7.8) 2 From Eq. (7.3), we have vm = imR v m i m or, = R 2 2 or, V = IR (7.9) Equation (7.9) gives the relation between ac current and ac voltage and is similar to that in the dc case. This shows the advantage of introducing the concept of rms values. In terms of rms values, the equation for power [Eq. (7.7)] and relation between current and voltage in ac circuits are essentially the same as those for the dc case. It is customary to measure and specify rms values for ac quantities. For example, the household line voltage of 220 V is an rms value with a peak voltage of vm = 2 V = (1.414)(220 V) = 311 V In fact, the I or rms current is the equivalent dc current that would produce the same average power loss as the alternating current. Equation (7.7) can also be written as P = V2 / R = I V (since V = I R) Example 7.1 A light bulb is rated at 100W for a 220 V supply. Find (a) the resistance of the bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. Solution (a) We are given P = 100 W and V = 220 V. The resistance of the bulb is 2 V 2 ( 220 V ) R = = = 484 Ω P 100 W (b) The peak voltage of the source is V 7.1 v m = 2V = 311 (c) Since, P = I V P 100 W I 0.454A EXAMPLE V 220 V Reprint 2025-26 Alternating Current

6.7 INDUCTANCE An electric current can be induced in a coil by flux change produced by another coil in its vicinity or flux change produced by the same coil. These two situations are described separately in the next two sub-sections. However, in both the cases, the flux through a coil is proportional to the current. That is, ΦB α I. Further, if the geometry of the coil does not vary with time then, dΦB d I ∝ d t d t For a closely wound coil of N turns, the same magnetic flux is linked with all the turns. When the flux ΦB through the coil changes, each turn contributes to the induced emf. Therefore, a term called flux linkage is used which is equal to NΦB for a closely wound coil and in such a case NΦB∝ I The constant of proportionality, in this relation, is called inductance. We shall see that inductance depends only on the geometry of the coil and intrinsic material properties. This aspect is akin to capacitance which for a parallel plate capacitor depends on the plate area and plate separation (geometry) and the dielectric constant K of the intervening medium (intrinsic material property). Inductance is a scalar quantity. It has the dimensions of [M L2 T–2 A–2] given by the dimensions of flux divided by the dimensions of current. The SI unit of inductance is henry and is denoted by H. It is named in honour of Joseph Henry who discovered electromagnetic induction in USA, independently of Faraday in England. 6.7.1 Mutual inductance Consider Fig. 6.12 which shows two long co-axial solenoids each of length l. We denote the radius of the inner solenoid S1 by r1 and the number of turns per unit length by n1. The corresponding quantities for the outer solenoid S2 are r2 and n2, respectively. Let N1 and N2 be the total number 165 of turns of coils S1 and S2, respectively. Reprint 2025-26 Physics When a current I2 is set up through S2, it in turn sets up a magnetic flux through S1. Let us denote it by Φ1. The corresponding flux linkage with solenoid S1 is N1 Φ1 = M 12 I 2 (6.7) M12 is called the mutual inductance of solenoid S1 with respect to solenoid S2. It is also referred to as the coefficient of mutual induction. For these simple co-axial solenoids it is possible to calculate M12. The magnetic field due to the current I2 in S2 is µ0n2I2. The resulting flux linkage with coil S1 is, πr12 N 1Φ1 = ( µ0n 2 I 2 ) n1l ) ( ) ( = µ0n 1n 2 πr12l I 2 (6.8) where n1l is the total number of turns in solenoid S1. FIGURE 6.12 Two long co-axial Thus, from Eq. (6.7) and Eq. (6.8), solenoids of same 2 M12 = µ0n1n2πr 1l (6.9) length l. Note that we neglected the edge effects and considered the magnetic field µ0n2I2 to be uniform throughout the length and width of the solenoid S2. This is a good approximation keeping in mind that the solenoid is long, implying l >> r2. We now consider the reverse case. A current I1 is passed through the solenoid S1 and the flux linkage with coil S2 is, N2Φ2 = M21 I1 (6.10) M21 is called the mutual inductance of solenoid S2 with respect to solenoid S1. The flux due to the current I1 in S1 can be assumed to be confined solely inside S1 since the solenoids are very long. Thus, flux linkage with solenoid S2 is πr12 N 2Φ2 = ( µ0n1 I1 ) n 2l ) ( ) ( where n2l is the total number of turns of S2. From Eq. (6.10), M21 = µ0n1n2πr 1l2 (6.11) Using Eq. (6.9) and Eq. (6.10), we get M12 = M21= M (say) (6.12) We have demonstrated this equality for long co-axial solenoids. However, the relation is far more general. Note that if the inner solenoid was much shorter than (and placed well inside) the outer solenoid, then we could still have calculated the flux linkage N1Φ1 because the inner solenoid is effectively immersed in a uniform magnetic field due to the outer solenoid. In this case, the calculation of M12 would be easy. However, it would be extremely difficult to calculate the flux linkage with the outer solenoid as the magnetic field due to the inner solenoid would vary across the length as well as cross section of the outer solenoid. Therefore, the calculation of M21 would also be extremely difficult in this case. The 166 equality M12=M21 is very useful in such situations. Reprint 2025-26 Electromagnetic Induction We explained the above example with air as the medium within the solenoids. Instead, if a medium of relative permeability µr had been present, the mutual inductance would be M =µr µ0 n1n2π r21 l It is also important to know that the mutual inductance of a pair of coils, solenoids, etc., depends on their separation as well as their relative orientation. Example 6.8 Two concentric circular coils, one of small radius r1 and the other of large radius r2, such that r1 << r2, are placed co-axially with centres coinciding. Obtain the mutual inductance of the arrangement. Solution Let a current I2 flow through the outer circular coil. The field at the centre of the coil is B2 = µ0I2 / 2r2. Since the other co-axially placed coil has a very small radius, B2 may be considered constant over its cross-sectional area. Hence, Φ1 = πr 1B22 µ0 π r12 = I 2 2r2 = M12 I2 Thus, µ0 πr12 M 12 = 2r2 From Eq. (6.12) µ0 πr12 M 12 = M 21 = 2 r2 Note that we calculated M12 from an approximate value of Φ1, assuming EXAMPLE the magnetic field B2 to be uniform over the area π r12. However, we can accept this value because r1 << r2. 6.8 Now, let us recollect Experiment 6.3 in Section 6.2. In that experiment, emf is induced in coil C1 wherever there was any change in current through coil C2. Let Φ1 be the flux through coil C1 (say of N1 turns) when current in coil C2 is I2. Then, from Eq. (6.7), we have N1Φ1 = MI2 For currents varrying with time, d d ( N 1Φ1 ) ( MI 2 ) = d t d t Since induced emf in coil C1 is given by d N 1Φ1 ) ( ε1 = – d t We get, d I 2 ε1 = – M d t 167 Reprint 2025-26 Physics It shows that varying current in a coil can induce emf in a neighbouring coil. The magnitude of the induced emf depends upon the rate of change of current and mutual inductance of the two coils. 6.7.2 Self-inductance In the previous sub-section, we considered the flux in one solenoid due to the current in the other. It is also possible that emf is induced in a single isolated coil due to change of flux through the coil by means of varying the current through the same coil. This phenomenon is called self-induction. In this case, flux linkage through a coil of N turns is proportional to the current through the coil and is expressed as NΦB ∝ I NΦB = L I (6.13) where constant of proportionality L is called self-inductance of the coil. It is also called the coefficient of self-induction of the coil. When the current is varied, the flux linked with the coil also changes and an emf is induced in the coil. Using Eq. (6.13), the induced emf is given by d ( NΦB ) ε = – dt d I ε = – L (6.14) d t Thus, the self-induced emf always opposes any change (increase or decrease) of current in the coil. It is possible to calculate the self-inductance for circuits with simple geometries. Let us calculate the self-inductance of a long solenoid of cross- sectional area A and length l, having n turns per unit length. The magnetic field due to a current I flowing in the solenoid is B = µ0 n I (neglecting edge effects, as before). The total flux linked with the solenoid is n I A ) NΦB = (nl )(µ0 )( = µ0n2 Al I where nl is the total number of turns. Thus, the self-inductance is, ΝΦΒ L = I = µ0n 2 Al (6.15) If we fill the inside of the solenoid with a material of relative permeability µr (for example soft iron, which has a high value of relative permeability), then, L = µr µ0 n 2 Al (6.16) The self-inductance of the coil depends on its geometry and on the permeability of the medium. The self-induced emf is also called the back emf as it opposes any 168 change in the current in a circuit. Physically, the self-inductance plays Reprint 2025-26 Electromagnetic Induction the role of inertia. It is the electromagnetic analogue of mass in mechanics. So, work needs to be done against the back emf (ε) in establishing the current. This work done is stored as magnetic potential energy. For the current I at an instant in a circuit, the rate of work done is d W = ε I d t If we ignore the resistive losses and consider only inductive effect, then using Eq. (6.14), d W d I = L I d t d t Total amount of work done in establishing the current I is I W = ∫ d W = ∫ L I d I 0 Thus, the energy required to build up the current I is, 1 2 W = LI (6.17) 2 This expression reminds us of mv 2/2 for the (mechanical) kinetic energy of a particle of mass m, and shows that L is analogous to m (i.e., L is electrical inertia and opposes growth and decay of current in the circuit). Consider the general case of currents flowing simultaneously in two nearby coils. The flux linked with one coil will be the sum of two fluxes which exist independently. Equation (6.7) would be modified into N1 Φ1 = M 11 I 1 + M 12 I 2 where M11 represents inductance due to the same coil. Therefore, using Faraday’s law, d I 1 d I 2 ε1 = − M 11 − M 12 d t d t M11 is the self-inductance and is written as L1. Therefore, d I 1 d I 2 ε1 = − L 1 − M 12 d t d t Example 6.9 (a) Obtain the expression for the magnetic energy stored in a solenoid in terms of magnetic field B, area A and length l of the solenoid. (b) How does this magnetic energy compare with the electrostatic energy stored in a capacitor? Solution (a) From Eq. (6.17), the magnetic energy is 1 2 U B = LI 2 2 EXAMPLE 1  B  L = ( since B = µ0 nI , for a solenoid ) 2  µ0 n  6.9 169 Reprint 2025-26 Physics 1 2  B  2 = ( µ0 n Al ) [from Eq. (6.15)] 2  µ0n  1 2 = B Al 2µ0 (b) The magnetic energy per unit volume is, U B u B = (where V is volume that contains flux) V U B = Al B 2 generator: = (6.18) 2µ0 ac We have already obtained the relation for the electrostatic energy on stored per unit volume in a parallel plate capacitor (refer to Chapter 2, Eq. 2.73), 1 2 u Ε = ε0 E (2.73) animation 2 6.9 In both the cases energy is proportional to the square of the field strength. Equations (6.18) and (2.73) have been derived for special cases: a solenoid and a parallel plate capacitor, respectively. But they Interactive http://micro.magnet.fsu.edu/electromag/java/generator/ac.html are general and valid for any region of space in which a magnetic field EXAMPLE or/and an electric field exist. 6.8 AC GENERATOR The phenomenon of electromagnetic induction has been technologically exploited in many ways. An exceptionally important application is the generation of alternating currents (ac). The modern ac generator with a typical output capacity of 100 MW is a highly evolved machine. In this section, we shall describe the basic principles behind this machine. The Yugoslav inventor Nicola Tesla is credited with the development of the machine. As was pointed out in Section 6.3, one method to induce an emf or current in a loop is through a change in the loop’s orientation or a change in its effective area. As the coil rotates in a magnetic field B, the effective area of the loop (the face perpendicular to the field) is A cos q, where q is the angle between A and B. This method of producing a FIGURE 6.13 AC Generator flux change is the principle of operation of a Reprint 2025-26 Electromagnetic Induction simple ac generator. An ac generator converts mechanical energy into electrical energy. The basic elements of an ac generator are shown in Fig. 6.13. It consists of a coil mounted on a rotor shaft. The axis of rotation of the coil is perpendicular to the direction of the magnetic field. The coil (called armature) is mechanically rotated in the uniform magnetic field by some external means. The rotation of the coil causes the magnetic flux through it to change, so an emf is induced in the coil. The ends of the coil are connected to an external circuit by means of slip rings and brushes. When the coil is rotated with a constant angular speed w, the angle q between the magnetic field vector B and the area vector A of the coil at any instant t is q = wt (assuming q = 0° at t = 0). As a result, the effective area of the coil exposed to the magnetic field lines changes with time, and from Eq. (6.1), the flux at any time t is FB = BA cos q = BA cos wt From Faraday’s law, the induced emf for the rotating coil of N turns is then, dΦB d ε= – N = – NBA (cos ωt ) dt d t Thus, the instantaneous value of the emf is ε= NBA ωsin ωt (6.19) where NBAw is the maximum value of the emf, which occurs when sin wt = ±1. If we denote NBAw as e0, then e = e0 sin wt (6.20) Since the value of the sine fuction varies between +1 and –1, the sign, or polarity of the emf changes with time. Note from Fig. 6.14 that the emf has its extremum value when q = 90° or q = 270°, as the change of flux is greatest at these points. The direction of the current changes periodically and therefore the current is called alternating current (ac). Since w = 2pn, Eq (6.20) can be written as e = e0sin 2p n t (6.21) where n is the frequency of revolution of the generator’s coil. Note that Eq. (6.20) and (6.21) give the instantaneous value of the emf and e varies between +e0 and –e0 periodically. We shall learn how to determine the time-averaged value for the alternating voltage and current in the next chapter. In commercial generators, the mechanical energy required for rotation of the armature is provided by water falling from a height, for example, from dams. These are called hydro-electric generators. Alternatively, water is heated to produce steam using coal or other sources. The steam at high pressure produces the rotation of the armature. These are called thermal generators. Instead of coal, if a nuclear fuel is used, we get nuclear power generators. Modern day generators produce electric power as high as 500 MW, i.e., one can light 171 Reprint 2025-26 Physics FIGURE 6.14 An alternating emf is generated by a loop of wire rotating in a magnetic field. up 5 million 100 W bulbs! In most generators, the coils are held stationary and it is the electromagnets which are rotated. The frequency of rotation is 50 Hz in India. In certain countries such as USA, it is 60 Hz. Example 6.10 Kamla peddles a stationary bicycle. The pedals of the bicycle are attached to a 100 turn coil of area 0.10 m2. The coil rotates at half a revolution per second and it is placed in a uniform magnetic field of 0.01 T perpendicular to the axis of rotation of the coil. What is the maximum voltage generated in the coil? Solution Here n = 0.5 Hz; N =100, A = 0.1 m2 and B = 0.01 T. Employing Eq. (6.19) e0 = NBA (2 p n) = 100 × 0.01 × 0.1 × 2 × 3.14 × 0.5 = 0.314 V 6.10 The maximum voltage is 0.314 V. We urge you to explore such alternative possibilities for power generation. EXAMPLE Reprint 2025-26 Electromagnetic Induction SUMMARY 1. The magnetic flux through a surface of area A placed in a uniform magnetic field B is defined as, FB = B.A = BA cos q where q is the angle between B and A. 2. Faraday’s laws of induction imply that the emf induced in a coil of N turns is directly related to the rate of change of flux through it, dΦB ε = − N d t Here FB is the flux linked with one turn of the coil. If the circuit is closed, a current I = e/R is set up in it, where R is the resistance of the circuit. 3. Lenz’s law states that the polarity of the induced emf is such that it tends to produce a current which opposes the change in magnetic flux that produces it. The negative sign in the expression for Faraday’s law indicates this fact. 4. When a metal rod of length l is placed normal to a uniform magnetic field B and moved with a velocity v perpendicular to the field, the induced emf (called motional emf) across its ends is e = Bl v 5. Inductance is the ratio of the flux-linkage to current. It is equal to NF/I. 6. A changing current in a coil (coil 2) can induce an emf in a nearby coil (coil 1). This relation is given by, d I 2 ε1 = − M 12 d t The quantity M12 is called mutual inductance of coil 1 with respect to coil 2. One can similarly define M21. There exists a general equality, M12 = M21 7. When a current in a coil changes, it induces a back emf in the same coil. The self-induced emf is given by, d I ε = − L d t L is the self-inductance of the coil. It is a measure of the inertia of the coil against the change of current through it. 8. The self-inductance of a long solenoid, the core of which consists of a magnetic material of relative permeability mr, is given by L = mr m0 n2 Al where A is the area of cross-section of the solenoid, l its length and n the number of turns per unit length. 9. In an ac generator, mechanical energy is converted to electrical energy by virtue of electromagnetic induction. If coil of N turn and area A is rotated at n revolutions per second in a uniform magnetic field B, then the motional emf produced is e = NBA (2pn) sin (2pnt) where we have assumed that at time t = 0 s, the coil is perpendicular to the field. 173 Reprint 2025-26 Physics Quantity Symbol Units Dimensions Equations Magnetic Flux FB Wb (weber) [M L2 T –2 A–1] FB = B i A EMF e V (volt) [M L2 T –3 A–1] e = − d( NΦB )/d t Mutual Inductance M H (henry) [M L2 T –2 A–2] e1 = − M 12 ( dI 2 /d t ) Self Inductance L H (henry) [M L2 T –2 A–2] ε = − L ( d I /d t ) POINTS TO PONDER 1. Electricity and magnetism are intimately related. In the early part of the nineteenth century, the experiments of Oersted, Ampere and others established that moving charges (currents) produce a magnetic field. Somewhat later, around 1830, the experiments of Faraday and Henry demonstrated that a moving magnet can induce electric current. 2. In a closed circuit, electric currents are induced so as to oppose the changing magnetic flux. It is as per the law of conservation of energy. However, in case of an open circuit, an emf is induced across its ends. How is it related to the flux change? 3. The motional emf discussed in Section 6.5 can be argued independently from Faraday’s law using the Lorentz force on moving charges. However, even if the charges are stationary [and the q (v × B) term of the Lorentz force is not operative], an emf is nevertheless induced in the presence of a time-varying magnetic field. Thus, moving charges in static field and static charges in a time-varying field seem to be symmetric situation for Faraday’s law. This gives a tantalising hint on the relevance of the principle of relativity for Faraday’s law. EXERCISES 6.1 Predict the direction of induced current in the situations described by the following Figs. 6.15(a) to (f ). Reprint 2025-26 Electromagnetic Induction FIGURE 6.15

4.5 MAGNETIC FIELD ON THE AXIS OF A CIRCULAR CURRENT LOOP In this section, we shall evaluate the magnetic field due to a circular coil along its axis. The evaluation entails summing up the effect of infinitesimal current elements (I dl) mentioned in the previous section. We assume that the current I is steady and that the evaluation is carried out in free space (i.e., vacuum). Fig. 4.9 depicts a circular loop carrying a steady current I. The loop is placed in the plane with its centre at the origin O and has a radius R. The x-axis is the axis of the loop. We wish to calculate the magnetic field at the point P on this axis. Let x be the distance of P from the centre O of the loop. Consider a conducting element dl of the loop. This is FIGURE 4.9 Magnetic field on the shown in Fig. 4.9. The magnitude dB of the magnetic axis of a current carrying circular field due to dl is given by the Biot-Savart law [Eq. 4.7(a)], loop of radius R. Shown are the µ0 I d l × r magnetic field dB (due to a line dB = 3 (4.8) element dl ) and its 4π r components along and Now r2 = x2 + R2 . Further, any element of the loop perpendicular to the axis. will be perpendicular to the displacement vector from the element to the axial point. For example, the element dl in Fig. 4.9 is in the plane, whereas, the displacement vector r from dl to the axial point P is in the plane. Hence |dl × r|=r dl. Thus, µ0 Idl d B = 4 π x 2 + R 2 (4.9) 115 ( ) Reprint 2025-26 Physics The direction of dB is shown in Fig. 4.9. It is perpendicular to the plane formed by dl and r. It has an x-component dBx and a component perpendicular to x-axis, dB⊥. When the components perpendicular to the x-axis are summed over, they cancel out and we obtain a null result. For example, the dB⊥ component due to dl is cancelled by the contribution due to the diametrically opposite dl element, shown in Fig. 4.9. Thus, only the x-component survives. The net contribution along x-direction can be obtained by integrating dBx = dB cos θ over the loop. For Fig. 4.9, R cosθ= 2 2 1/2 (4.10) ( x + R ) From Eqs. (4.9) and (4.10), µ0 Idl R d B x = 4 π x 2 + R 2 3/2 ( ) The summation of elements dl over the loop yields 2πR, the circumference of the loop. Thus, the magnetic field at P due to entire circular loop is 2 µ 0 I R ˆ i B = B x ˆi = 3/2 (4.11) 2 2 2 x + R ( ) As a special case of the above result, we may obtain the field at the centre of the loop. Here x = 0, and we obtain, B 0 = µ0 I ˆi (4.12) 2 R The magnetic field lines due to a circular wire form closed loops and are shown in Fig. 4.10. The direction of the magnetic field is given by (another) right-hand thumb rule stated below: Curl the palm of your right hand around the circular wire with the fingers pointing in the direction of the current. The right-hand thumb gives the direction of the magnetic field. FIGURE 4.10 The magnetic field lines for a current loop. The direction of the field is given by the right-hand thumb rule described in the text. The upper side of the loop may be thought of as the north pole and the lower 116 side as the south pole of a magnet. Reprint 2025-26 Moving Charges and Magnetism Example 4.5 A straight wire carrying a current of 12 A is bent into a semi-circular arc of radius 2.0 cm as shown in Fig. 4.11(a). Consider the magnetic field B at the centre of the arc. (a) What is the magnetic field due to the straight segments? (b) In what way the contribution to B from the semicircle differs from that of a circular loop and in what way does it resemble? (c) Would your answer be different if the wire were bent into a semi-circular arc of the same radius but in the opposite way as shown in Fig. 4.11(b)? FIGURE 4.11 Solution (a) dl and r for each element of the straight segments are parallel. Therefore, dl × r = 0. Straight segments do not contribute to |B|. (b) For all segments of the semicircular arc, dl × r are all parallel to each other (into the plane of the paper). All such contributions add up in magnitude. Hence direction of B for a semicircular arc is given by the right-hand rule and magnitude is half that of a circular loop. Thus B is 1.9 × 10–4 T normal to the plane of the EXAMPLE paper going into it. (c) Same magnitude of B but opposite in direction to that in (b). 4.5 Example 4.6 Consider a tightly wound 100 turn coil of radius 10 cm, carrying a current of 1 A. What is the magnitude of the magnetic field at the centre of the coil? Solution Since the coil is tightly wound, we may take each circular element to have the same radius R = 10 cm = 0.1 m. The number of turns N = 100. The magnitude of the magnetic field is, EXAMPLE –7 NI µ 4 π × 10 × 10 2 × 1 0 B = = –1 = 2 π × 10−4 = 6.28 × 10 −4 T 4.6 R 2 10 2 ×